<HTML>

<HEAD>

<TITLE>Special Relativity: What Time is it? </TITLE>

<META NAME="GENERATOR" CONTENT="Internet Assistant for Microsoft Word 2.0z">
</HEAD>

<BODY bgcolor="#ffffff">

<H2>Special Relativity: What Time is it? </H2>

<P>
<I>Michael Fowler</I>
<P>
<I>Physics Department, UVa. </I>
<P>
<A HREF="lecturelist.html">Index of Lectures and Overview of the Course</A><BR>
<A HREF="spec_rel.html" >Link to Previous Lecture</A>
<H3>All Inertial Frames Look the Same </H3>

<P>
Einstein's Theory of Special Relativity, discussed in the last
lecture, may be summarized as follows:
<P>
<I>The Laws of Physics are the same in any Inertial Frame of Reference.
(Such frames move at steady velocities with respect to each other.)
These Laws include not only Newton's Laws of motion, but also
the more recently discovered Maxwell's Equations describing electric
and magnetic fields, which predict that light always travels at
a speed c, equal to about 186,300 miles per second. <B>It follows
that any measurement of the speed of any flash of light by an
observer in an inertial frame will give the answer c.</B></I>
<P>
We have already noted one counter-intuitive consequence of this,
that two different observers moving relative to each other, each
measuring the speed of the <I>same</I> blob of light relative
to himself, will<I> both</I> get <I>c</I>, even if their relative
motion is in the same direction as the motion of the blob of light.
<H3>A Simple but Reliable Clock </H3>

<P>
We mentioned earlier that each of our (inertial) frames of reference
is calibrated (had marks at regular intervals along the walls)
to measure distances, and has a clock to measure time. Let us
now get more specific about the clock---we want one that is easy
to understand in any frame of reference. Instead of a pendulum
swinging back and forth, which wouldn't work away from the earth's
surface anyway, we have a blip of light bouncing back and forth
between two mirrors facing each other. We call this device a <I>light
clock</I>. To really use it as a timing device we need some way
to count the bounces, so we position a photocell at the upper
mirror, so that it catches the edge of the blip of light. The
photocell clicks when the light hits it, and this regular series
of clicks drives the clock hand around, just as for an ordinary
clock. Of course, driving the photocell will eventually use up
the blip of light, so we also need some provision to reinforce
the blip occasionally, such as a strobe light set to flash just
as it passes and thus add to the intensity of the light. Admittedly,
this may not be an easy way to build a clock, but the basic idea
is simple. 
<P>
<IMG SRC="photclk1.gif" ALIGN="BOTTOM">
<P>
It's easy to figure out how frequently our light clock clicks.
If the two mirrors are a distance <I>w</I> apart, the round trip
distance for the blip from the photocell mirror to the other mirror
and back is 2<I>w</I>. Since we know the blip always travels at
<I>c</I>, we find the round trip time to be 2<I>w/c</I>, so this
is the time between clicks. This isn't a very long time for a
reasonable sized clock! The crystal in a quartz watch &quot;clicks
&quot; of the order of 10,000 times a second. That would correspond
to mirrors about nine miles apart, so we need our clock to click
about 1,000 times faster than that to get to a reasonable size.
Anyway, let us assume that such purely technical problems have
been solved.
<H3>Looking at Somebody Else's Clock </H3>

<P>
Let us now consider two observers, Jack and Jill, each equipped
with a calibrated inertial frame of reference, and a light clock.
To be specific, imagine Jack standing on the ground with his light
clock next to a straight railroad line, while Jill and her clock
are on a large flatbed railroad wagon which is moving down the
track at a constant speed <I>v</I>. Jack now decides to check
Jill's light clock against his own. He knows the time for his
clock is 2<I>w/c</I> between clicks. Imagine it to be a slightly
misty day, so with binoculars he can actually see the blip of
light bouncing between the mirrors of Jill's clock. How long does
he think that blip takes to make a round trip? The one thing he's
sure of is that it must be moving at <I>c </I>= 186,300 miles
per second, relative to him -- that's what Einstein tells him.
So to find the round trip time, all he needs is the round trip
distance. This will <I>not </I>be 2<I>w</I>, because the mirrors
are on the flatbed wagon moving down the track, so, relative to
Jack on the ground, when the blip gets back to the top mirror,
that mirror has moved down the track some since the blip left,
so the blip actually follows a zigzag path as seen from the ground.

<P>
<IMG SRC="photclk0.gif" ALIGN="BOTTOM">
<P>
Suppose now the blip in Jill's clock on the moving flatbed wagon
takes time <I>t</I> to get from the bottom mirror to the top mirror
as measured by Jack standing by the track. Then the length of
the &quot;zig&quot; from the bottom mirror to the top mirror is
necessarily <I>ct</I>, since that is the distance covered by any
blip of light in time <I>t</I>. Meanwhile, the wagon has moved
down the track a distance <I>vt</I>, where <I>v</I> is the speed
of the wagon. This should begin to look familiar---it is precisely
the same as the problem of the swimmer who swims at speed c relative
to the water crossing a river flowing at <I>v</I>! We have again
a right-angled triangle with hypotenuse <I>ct</I>, and shorter
sides <I>vt</I> and <I>w</I>. 
<P>
From Pythagoras, then, 
<P>
<CENTER><I>c</I>&#178;<I>t</I>&#178; = <I>v</I>&#178;<I>t</I>&#178;
+ <I>w</I>&#178;</CENTER>
<P>
so
<P>
<CENTER><I>t</I>&#178;(<I>c</I>&#178; - <I>v</I>&#178;) = <I>w</I>&#178;</CENTER>
<P>
or
<P>
<CENTER> <I>t</I>&#178;(1 - <I>v</I>&#178;/<I>c</I>&#178;) = <I>w</I>&#178;/<I>c</I>&#178;
<BR>
</CENTER>
<P>
and, taking the square root of each side, then doubling to get
the round trip time, we conclude that Jack sees the time between
clicks for Jill's clock to be:
<P>
<CENTER>2<I>w/c</I>&times;1/sqrt(1-<I>v</I>&#178;<I>/c</I>&#178;)</CENTER>
<P>
This means that Jack sees Jill's light clock to be going slow---a
longer time between clicks---compared to his own identical clock.
Obviously, the effect is not dramatic at real railroad speeds.
The correction factor is sqrt(1-<I>v</I>&#178;/<I>c</I>&#178;),
which differs from 1 by about one part in a trillion even for
a bullet train! Nevertheless, the effect is real and can be measured,
as we shall discuss later.
<P>
It is important to realize that the only reason we chose a <I>light</I>
clock, as opposed to some other kind of clock, is that its motion
is very easy to analyze from a different frame. Jill could have
a collection of clocks on the wagon, and would synchronize them
all. For example, she could hang her wristwatch right next to
the face of the light clock, and observe them together to be sure
they always showed the same time. Remember, in her frame her light
clock clicks every 2<I>w/c</I> seconds, as it is designed to do.
Observing this scene from his position beside the track, Jack
will see the synchronized light clock and wristwatch next to each
other, and, of course, note that the wristwatch is <I>also</I>
running slow by the factor sqrt(1-<I>v</I>&#178;<I>/c</I>&#178;).
In fact, <I>all</I> her clocks, including her pulse, are slowed
down by this factor according to Jack. Jill is aging more slowly
because she's moving!
<P>
But this isn't the whole story -- we must now turn everything
around and look at it from Jill's point of view. Her inertial
frame of reference is just as good as Jack's. She sees his light
clock to be moving at speed <I>v</I> (backwards) so from her point
of view <I>his</I> light blip takes the longer zigzag path, which
means <I>his clock runs slower than hers.</I> That is to say,
each of them will see the other to have slower clocks, and be
aging more slowly. This phenomenon is called <I>time dilation</I>.
It has been verified in recent years by flying very accurate clocks
around the world on jetliners and finding they register less time,
by the predicted amount, than identical clocks left on the ground.
Time dilation is very easy to observe in elementary particle physics,
as we shall discuss in the next section.
<H3>Fitzgerald Contraction </H3>

<P>
Consider now the following puzzle: suppose Jill's clock is equipped
with a device that stamps a notch on the track once a second.
How far apart are the notches? From Jill's point of view, this
is pretty easy to answer. She sees the track passing under the
wagon at <I>v </I>feet per second, say, to choose convenient units,
so the notches will of course be <I>v</I> feet apart. But Jack
sees things differently. He sees Jill's clocks to be running slow,
so he will see the notches to be stamped on the track at intervals
of 1/sqrt(1-<I>v</I>&#178;/<I>c</I>&#178;) seconds (so for a relativistic
train going at <I>v</I> = 0.8<I>c</I>, the notches are stamped
at intervals of 5/3 = 1.67 seconds). Since Jack agrees with Jill
that the relative speed of the wagon and the track is <I>v</I>,
he will assert the notches are not <I>v</I> feet apart, but <I>v</I>/sqrt(1-<I>v</I>&#178;/<I>c</I>&#178;)
feet apart, a greater distance. Who is right? It turns out that
Jack is right, because the notches are in his frame of reference,
so he can wander over to them with a tape measure or whatever,
and check the distance. This implies that as a result of her motion,
Jill observes the notches to be closer together by a factor sqrt(1-<I>v</I>&#178;/<I>c</I>&#178;)
than they would be at rest. This is called the Fitzgerald contraction,
and applies not just to the notches, but also to the track and
to Jack -- everything looks somewhat squashed in the direction
of motion!
<H3>Experimental Evidence for Time Dilation: Dying Muons </H3>

<P>
The first clear example of time dilation was provided over fifty
years ago by an experiment detecting <I>muons</I>. These particles
are produced at the outer edge of our atmosphere by incoming cosmic
rays hitting the first traces of air. They are unstable particles,
with a &quot;half-life&quot; of 1.5 microseconds (1.5 millionths
of a second), which means that if at a given time you have 100
of them, 1.5 microseconds later you will have about 50, 1.5 microseconds
after that 25, and so on. Anyway, they are constantly being produced
many miles up, and there is a constant rain of them towards the
surface of the earth, moving at very close to the speed of light.
In 1941, a detector placed near the top of Mount Washington (at
6000 feet above sea level) measured about 570 muons per hour coming
in. Now these muons are raining down from above, but dying as
they fall, so if we move the detector to a lower altitude we expect
it to detect fewer muons because a fraction of those that came
down past the 6000 foot level will die before they get to a lower
altitude detector. Approximating their speed by that of light,
they are raining down at 186,300 miles per second, which turns
out to be, conveniently, about 1,000 feet per microsecond. Thus
they should reach the 4500 foot level 1.5 microseconds after passing
the 6000 foot level, so, if half of them die off in 1.5 microseconds,
as claimed above, we should only expect to register about 570/2
= 285 per hour with the same detector at this level. Dropping
another 1500 feet, to the 3000 foot level, we expect about 280/2
= 140 per hour, at 1500 feet about 70 per hour, and at ground
level about 35 per hour. (We have rounded off some figures a bit,
but this is reasonably close to the expected value.) 
<P>
To summarize: given the known rate at which these raining-down
unstable muons decay, and given that 570 per hour hit a detector
near the top of Mount Washington, we only expect about 35 per
hour to survive down to sea level. In fact, when the detector
was brought down to sea level, it detected about 400 per hour!
How did they survive? The reason they didn't decay is that <I>in
their frame of reference, much less time had passed</I>. Their
actual speed is about 0.994<I>c</I>, corresponding to a time dilation
factor of about 9, so in the 6 microsecond trip from the top of
Mount Washington to sea level, their clocks register only 6/9
= 0.67 microseconds. In this period of time, only about one-quarter
of them decay.
<P>
What does this look like from the muon's point of view? How do
they manage to get so far in so little time? To them, Mount Washington
and the earth's surface are approaching at 0.994c, or about 1,000
feet per microsecond. But in the 0.67 seconds it takes them to
get to sea level, it would seem that to them sea level could only
get 670 feet closer, so how could they travel the whole 6000 feet
from the top of Mount Washington? The answer is the Fitzgerald
contraction---to them Mount Washington is squashed in a vertical
direction (the direction of motion) by a factor of 1/sqrt(1-<I>v</I>&#178;/<I>c</I>&#178;),
the same as the time dilation factor, which for the muons is 9.
So, to the muons, Mount Washington is only 670 feet high---this
is why they can get down it so fast!
<P>
<A HREF="synchronizing.html">Link to Next Lecture</A><BR><P>
<A HREF="lecturelist.html">Index of Lectures and Overview of the Course</A><BR>
<P>
Copyright &copy;1996 Michael Fowler 
</BODY>

</HTML>
